If the straight lines `ax+by+p=0` and `x cos alpha+ysin alpha-p=0` enclose an angle `pi//4` between them and meet the straight line `x sin alpha-y cos alpha=0` in the same point, then the value of `a^(2)+b^(2)` is equal to ………..

Answer

To solve the problem, we need to find the value of \( a^2 + b^2 \) given the conditions about the lines.

### Step 1: Identify the slopes of the lines
The first line is given by the equation:
\[ ax + by + p = 0 \]
The slope \( m_1 \) of the first line can be calculated as:
\[ m_1 = -\frac{a}{b} \]

The second line is given by:
\[ x \cos \alpha + y \sin \alpha - p = 0 \]
The slope \( m_2 \) of the second line is:
\[ m_2 = -\frac{\cos \alpha}{\sin \alpha} \]

The third line is given by:
\[ x \sin \alpha - y \cos \alpha = 0 \]
The slope \( m_3 \) of the third line is:
\[ m_3 = \frac{\sin \alpha}{\cos \alpha} \]

### Step 2: Use the angle condition
The problem states that the angle between the first line and the second line is \( \frac{\pi}{4} \). The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Setting \( \theta = \frac{\pi}{4} \), we have \( \tan \frac{\pi}{4} = 1 \). Thus:
\[ 1 = \left| \frac{-\frac{a}{b} + \frac{\cos \alpha}{\sin \alpha}}{1 - \frac{a}{b} \cdot \frac{\cos \alpha}{\sin \alpha}} \right| \]

### Step 3: Simplify the equation
Cross-multiplying gives:
\[ 1 - \frac{a \cos \alpha}{b \sin \alpha} = -\frac{a}{b} + \frac{\cos \alpha}{\sin \alpha} \]
Rearranging terms leads to:
\[ 1 + \frac{a}{b} = \frac{\cos \alpha}{\sin \alpha} + \frac{a \cos \alpha}{b \sin \alpha} \]

### Step 4: Analyze the intersection point
The lines \( L_2 \) and \( L_3 \) are also given to meet at the same point. The slopes \( m_2 \) and \( m_3 \) must also satisfy the angle condition:
\[ \tan \theta = \left| \frac{m_2 - m_3}{1 + m_2 m_3} \right| = 1 \]

### Step 5: Set up the equations
Using the slopes:
\[ 1 = \left| \frac{-\frac{\cos \alpha}{\sin \alpha} - \frac{\sin \alpha}{\cos \alpha}}{1 - \left(-\frac{\cos \alpha}{\sin \alpha}\right) \left(\frac{\sin \alpha}{\cos \alpha}\right)} \right| \]

### Step 6: Solve for \( a^2 + b^2 \)
After simplifying and equating the conditions from both angle equations, we can derive:
\[ a^2 + b^2 = 0 \]

### Conclusion
Thus, the value of \( a^2 + b^2 \) is:
\[ \boxed{1} \]